Question

Find a particular solution of y"+2y'+2y=e^(-x)(8cosx-6sinx).

Answer 1

\[y _{p}=xe ^{-x}(Acosx+Bsinx)\]

Answer 2

Is my initial guess right above?

Answer 3

you've taken the complementary solution and just tines'd it by x. doesn't work that way.

you could, having worked the complementary solution, then go through the Variation of Params route.

Answer 4

to be sure, there are other ways to go about it

eg you can re-write the RHS as \(10 e^{-x} (\frac{4}{5} \cos x - \frac{3}{5} \sin x ) = 10 e^{-x} \mathcal{Re } \{ e^{i(x + \alpha)}\}\) and work it like that, ie solve \(z_p = x_p + i y_p\), but looks like an awful mess

or you can guess the solution, aka Method of Undetermined Coefficients, but I have no idea OTTOMF what a good guess would be

Var of Params should just work, described here:

http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx

pretty sure someone else will think of other ways too :-)

Answer 5

\(e^{-x}(A\cos(x)-B\sin(x))\) looks like a good guess to me.

Answer 6

\(\color{blue}{\text{Originally Posted by}}\) @Idealist10
\[y _{p}=xe ^{-x}(Acosx+Bsinx)\]
\(\color{blue}{\text{End of Quote}}\)

i'd say the same thing as them,
\[y _{p}=e ^{-x}(Acosx+Bsinx)\]

is better

Answer 7

\[y _{p}=e ^{-x}(Acosx+Bsinx)\] doesn't work. I tried.

Answer 8

then you might have tried
\[y _{p}=xe ^{-x}(Acosx+Bsinx)\]
too? that should work ! if not, then even idk :P

Answer 9

The guess \(y_p=xe^{-x}\left(A\cos(x)+B\sin(x)\right)\) works. Try again.